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Sorryforgot to upload problem

  1. Nov 15, 2005 #1
    in the attachment there is a problem related to techniques of integration, i fully understand everything until near the end of the problem they answer says

    (1/6)o - (1/12)sin2o + C = (1/6)o - (1/6)sino coso + C

    and then they convert the sinocoso to sec-1

    this part is really confusing...can you help me plz...

    thank you
     

    Attached Files:

  2. jcsd
  3. Nov 16, 2005 #2

    benorin

    User Avatar
    Homework Helper

    Recall that [tex]\sin(2o)=2\sin (o)\cos (o)[/tex], this is the first one.

    Note that, from the triangle, we have [tex]\sec (\theta ) = \frac{x}{3} \Rightarrow \theta = \sec ^{-1} \left( \frac{x}{3}\right) [/tex]
    and, also from the triangle, we have [tex]\cos (\theta ) = \frac{3}{x}[/tex]

    and [tex]\sin (\theta ) = \mbox{ what? }[/tex], and the required result follows.
     
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